Question 75236
If 10% of the rivets produced by a machine are defective, find the probability that out of 5 randomly chosen rivets:
The problem is "binomial" because the rivets are defective or not defective.
P(defective)=0.10; P(not defective)=0.90
(i) P(one or more will be defective) = 1 - P(none are defective)
1-(0.90)^5 
= 0.40951
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If you use a TI calculator it is "1 - binompdf(5,0.1,0)"
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(ii)P(at most two will be defective)
=P(x=0)+P(x=1)+P(x=2)
=binomcdf(5,0.1,2)=0.99144
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Cheers,
Stan H.